[Algorithm] Finding Prime numbers - Sieve of Eratosthenes

【[Algorithm] Finding Prime numbers - Sieve of Eratosthenes】的更多相关文章

[Algorithm] Finding Prime numbers - Sieve of Eratosthenes

Given a number N, the output should be the all the prime numbers which is less than N. The solution is calledSieve of Eratosthenes: First of all, we assume all the number from 2 to N are prime number (0 & 1 is not Prime number). According to the Prim…

CSU 2018年12月月赛 F(2218): Finding prime numbers

Description xrdog has a number set. There are 95 numbers in this set. They all have something in common and are all positive integers less than 500. The naughty xiao zhuan has deleted one of the numbers in the set. Can you find the number that xiao z…

algorithm@ Sieve of Eratosthenes (素数筛选算法) & Related Problem (Return two prime numbers )

Sieve of Eratosthenes (素数筛选算法) Given a number n, print all primes smaller than or equal to n. It is also given that n is a small number. For example, if n is 10, the output should be “2, 3, 5, 7″. If n is 20, the output should be “2, 3, 5, 7, 11, 13,…

使用埃拉托色尼筛选法(the Sieve of Eratosthenes)在一定范围内求素数及反素数(Emirp)

Programming 1.3 In this problem, you'll be asked to find all the prime numbers from 1 to 1000. Prime numbers are used in allkinds of circumstances, particularly in fields such as cryptography, hashing among many others. Any method w ill be sufficient…

poj 2379 Sum of Consecutive Prime Numbers

Sum of Consecutive Prime Numbers Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 24498 Accepted: 13326 Description Some positive i…

POJ 2739 Sum of Consecutive Prime Numbers（尺取法）

题目链接: 传送门 Sum of Consecutive Prime Numbers Time Limit: 1000MS Memory Limit: 65536K Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations does a given positive intege…

[原]素数筛法【Sieve Of Eratosthenes + Sieve Of Euler】

拖了有段时间,今天来总结下两个常用的素数筛法: 1.sieve of Eratosthenes[埃氏筛法] 这是最简单朴素的素数筛法了,根据wikipedia,时间复杂度为 ,空间复杂度为O(n). 算法思想:先假定所有的数都是素数,然后从最小的素数2出发,把素数的所有倍数筛出去.又因为一个数的质因数都是成对出现的,比如100 = 1*100 = 2*50 = .....= 10*10,所以筛素数时只用筛到 n的开平方就行了. 伪代码如下: 对于任意的范围n, 设bool prime[ ],初始…